Quartic Equation

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A quartic equation is an algebraic equation of the form

$ax^4 + bx^3 + cx^2 + dx + e = 0.$

These types of equations are extremely hard to solve; however, there are very clever methods for solving them by bringing it down to a cubic. I am going to list the simplest of the five.

Solving Quartic Equations

Look in the "TLDR" section for the final result of each step.

Bringing it down to a depressed quartic

Start with the equation $ax^4 + bx^3 + cx^2 + dx + e = 0.$ Divide both sides by a: $x^4 + \frac{b}{a}x^3 + \frac{c}{a}x^2 + \frac{d}{a}x + \frac{e}{a} = 0$ Now, convert to a depressed quartic by substituting $x = y - \frac{b}{4a}$. You now have:

$\left(y - \frac{b}{4a}\right)^4 + \frac{b}{a}\left(y - \frac{b}{4a}\right)^3 + \frac{c}{a}\left(y - \frac{b}{4a}\right)^2 + \frac{d}{a}\left(y - \frac{b}{4a}\right) + \frac{e}{a} = 0$

$y^4 - \left(\frac{b}{a}\right)y^3 + \left(\frac{3b^2}{8a^2}\right)y^2 - \left(\frac{b^3}{16a^3}\right)y + \left(\frac{b^4}{256a^4}\right) + \left(\frac{b}{a}\right)y^3 - \left(\frac{6b^2}{8a^2}\right)y^2 + \left(\frac{3b^3}{16a^3}\right)y - \left(\frac{4b^4}{256a^4}\right) + \left(\frac{8ac}{8a^2}\right)y^2 - \left(\frac{8abc}{16a^3}\right)y$ $+ \left(\frac{16ab^2c}{256a^4}\right) + \left(\frac{16a^2d}{16a^3}\right)y - \left(\frac{64a^2bd}{256a^4}\right) + \left(\frac{256a^3e}{256a^4}\right) = 0$

$y^4 + \left(\frac{8ac - 3b^2}{8a^2}\right)y^2 + \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)y + \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right) = 0$

Now you have a depressed quartic: $y^4 + py^2 + qy + r = 0$ where $p = \left(\frac{8ac - 3b^2}{8a^2}\right)$, $q = \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)$ and $r = \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right)$.

Ferrari's Solution

Coming soon!

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